Question

If x - $\frac{1}{x}$ = 8, find the value of $x^{2}$ + $\frac{1}{x^2}$

Answer 1

        $\text{It is given that }x+\dfrac{1}{x} = 8$

        $\text{We must find }x^2 + \dfrac{1}{x^2}$

$\implies\: \text{Squaring on both sides}$

$\implies \: \text{\huge{(}}x + \dfrac{1}{x}\text{\huge{)}}^2 = (8)^2$

        $\text{We know that }(a+b)^2 = a^2 + b^2 + 2ab$

$\implies \: x^2 + \dfrac{1}{x^2} + 2(x)\text{\huge{(}}\dfrac{1}{x}\text{\huge{)}} = 64$

$\implies \: x^2 + \dfrac{1}{x^2} +2 = 64$

$\implies \: \boxed{x^2 + \dfrac{1}{x^2} = 62}$